Mathematics - Continuity - Differentiability Question with Solution | TestHub

Given : $f(x)=\{\begin{array}{ll}{a}_n+\sin \pi x,& x\in [2n,2n+1]\\ b_n+\cos \pi x,& x\in (2n-1,2n)\end{array}$ $\because f$ is continuous for all $n$ $\therefore$ At $x=2n,\mathrm{LHL}=\mathrm{RHL}=f(2n)$ $\Rightarrow b_n+\cos 2\pi n=a_n+\sin 2\pi n=a_n+\sin 2\pi n$ $\Rightarrow b_n+1=a_n\Rightarrow a_n-b_n=1,\therefore$ option (b) is correct. Also at $x=2n+1,\mathrm{LHL}=\mathrm{RHL}=f(2n+1)$ $\Rightarrow {\lim }_{h\to 0}{a}_n+\sin \pi (2n+1-h)$ $={\lim }_{h\to 0}{b}_{n+1}+\cos \pi (2n+1-h)=a_n+\sin (2n+1)\pi$ $\Rightarrow a_n=b_{n+1}-1=a_n\Rightarrow a_n-b_{n+1}=-1$ $\therefore$ option (c) is incorrect. $\Rightarrow a_{n-1}-b_n=-1,\therefore$ option (d) is correct.